SCEA131 September   2023 TXB0101 , TXB0102 , TXB0104 , TXB0104-Q1 , TXB0106 , TXB0106-Q1 , TXB0108 , TXB0302 , TXB0304 , TXS0101 , TXS0102 , TXS0102-Q1 , TXS0104E , TXS0104E-Q1 , TXS0108E , TXS0108E-Q1

 

  1.   1
  2.   Abstract
  3.   Trademarks
  4. 1Introduction
  5. 2One-Shot Performance
  6. 3Design Considerations
    1. 3.1 PCB Layout Design with the One-Shots
    2. 3.2 Effects of External Resistors on One-Shots
    3. 3.3 Effects of Slow Inputs on the One-Shots
    4. 3.4 Effects of Speed and Loading on One-Shots
    5. 3.5 Effects of Temperature on One-Shots
  7. 4Summary
  8. 5References

PCB Layout Design with the One-Shots

To help minimize noise with switching transients and further avoid false triggers from ringing / reflections while interacting with the one-shots, bypass capacitors should be used on both power supplies, VCCA and VCCB. Careful considerations should be made for short path bypass capacitors, as close as possible to both VCC supplies and GND while designing printed circuit boards (PCB) with devices with one-shots.

PCB signal trace-lengths must be kept short enough so that the round-trip delay of any reflection is less than the one-shot duration, typically 10ns to 30 ns. With minimal parasitic capacitance and inductance from shorter traces, reflections encounter low impedance at the source driver further avoiding false triggering of the one-shot for an abnormal output pulse width, output signal oscillations, or other adverse system-level effects.

The following equation for the board effective relative permittivity can be used to determine the maximum trace length to help remain within the design's stipulated delay:

Propagation Velocity, V= c/√(εr ).

where c is the speed of light and the wavelength of the signal, λ = V / bandwidth.

For example, typical bandwidth for logic devices are at 300 MHz and the typical effective permittivity of a PCB εr, around 4.

Signal integrity issues may be observed with trace lengths > 1/8th of the signal’s wavelength, (λ/8).

Hence, V = 3×10^8 (m/s) /√4= 15×10^7m/s

λ = 15×10^7 (m/s) / 300×10^6 (Hz) = 0.5 m

And, λ/8 = 0.625 m. For example, approximately 6 cm or approximately 2.5 inches.